Question: $\text D = \left[\begin{array}{rr}-1 & 0 \\ 2 & -2 \\ 1 & 3\end{array}\right]$ and $\text F = \left[\begin{array}{rr}0 & 2 \\ 3 & -2\end{array}\right]$ Let $\text {H = DF}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{D}$ and the first column of $\text{F}$. $ \text {H}=\left[\begin{array}{rr}{-1} & {0} \\ 2 & -2 \\ 1 & 3\end{array}\right]\left[\begin{array}{rr} {0} & 2 \\ {3} & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(-1,0)\cdot(0,3)\\\\ &=-1 \cdot 0 + 0\cdot 3\\\\ &=0 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 \cdot 2 + 0\cdot -2 = -2$ (Choice B) B $2 \cdot 0 - 2\cdot 3 = -6$ (Choice C) C $2 \cdot 3 - 2\cdot -2 = 10$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}0 & -2 \\ -6 & 8 \\ 9 & -4\end{array}\right]$